Is it true that $(\kappa = 0,1$ or infinite$)$ $\Leftrightarrow (\kappa^\mu = \kappa$ for all $\mu < \kappa)$?

Question

Clearly the statement holds for all cardinals $\kappa$ up to $\aleph_0$ - it is well known that $|\mathbb{N}^n| = |\mathbb{N}|$ for all $n \in \mathbb{N}$. Assuming the generalised continuum hypothesis, it seems that one may be able to prove such a thing inductively. However, does this hold without assuming $\mathsf{GCH}$?

Answer 1

By König's theorem you have $$\kappa^{\operatorname{cf}(\kappa)}>\kappa$$ for any $\kappa \ge \aleph_0$.

In particular, if you have $\mu=\operatorname{cf}(\kappa)<\kappa$, then you get $\kappa^\mu>\kappa$, which gives a counterexample to the claim from the question.

The inequality $\operatorname{cf}(\kappa)<\kappa$ is true for any singular cardinal $\kappa$; for example, for $\kappa=\aleph_\omega$ you have $\operatorname{cf}(\aleph_\omega)=\aleph_0$.

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Under GCH you can still prove that the claim is true for regular $\kappa$. Here is Corollary 1.7.5 from Holz, Michael; Steffens, Karsten; Weitz, Edmund: Introduction to Cardinal Arithmetic; Birkhäuser 1999.

Corollary 1.7.5. Assume (GCH). Then for all cardinals $\kappa\ge2$ and $\lambda\ge\omega$, $$\kappa^\lambda = \begin{cases} \kappa & \text{if }\lambda<\operatorname{cf}(\kappa), \\ \kappa^+ & \text{if }\operatorname{cf}(\kappa)\le\lambda<\kappa, \\ \lambda^+ & \text{if }\kappa\le\lambda. \end{cases} $$

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Without GCH, the equality can fail for regular cardinals. For example, in the Cohen's model for ZFC+$\neg$CH we have $2^{\aleph_0}=2^{\aleph_1}=\aleph_2$. This gives us $\aleph_1^{\aleph_0}=\aleph_2>\aleph_1$. (In fact, whenever we have $2^\mu=2^\kappa$, then $\kappa^\mu\ge 2^\mu=2^\kappa > \kappa$.)